Now say you’ve just solved a quadratic equation and you’ve got two answers, t = 5, –1. For this particular problem pretend that the solutions represent how long in seconds it takes a rocket to reach a certain height. The solution t = 5 is fine, but what does the solution t = –1 mean? This suggests that the rocket reaches a certain height before it is launched!
Obviously this cannot be true, so only the solution t = 5 should be used. This can happen quite often – you must pick the ‘sensible’ solutions. (t = –1 is a mathematical solution that doesn’t make any sense when you use it in a real life application).
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Suppose a rocket launches from a tower. Its height
above the ground after t seconds is |
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Solution |
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First of all, the formula looks like part of a quadratic equation, without the “= 0” at the end. The question is asking how long it is before the rocket hits the ground. This is the same as asking when will the rocket’s height above the ground equal 0. One more point about the height equation – at time t = 0, just before launch, the height is: This means the rocket starts 5 m above the ground. It’s probably mounted on a launching pad or something. Now we want to find out what ‘t’ equals when the height above the ground is zero – when it has come back to earth. We can do this by setting the equation for height to equal 0: Now we do have a quadratic equation! Let’s try solving it using factorisation: Try –t and t: Possible pairs: 5 and 1, –5 and –1. Try putting in 5 and 1: Multiply out and check: This isn’t the same equation – perhaps try swapping the 1 and 5 around: Now that we have the correct equation, let it equal 0: From this you can easily see that t = 5 and t = –1 are solutions (one of the brackets must equal 0). Now we have to see whether all the answers are ‘reasonable’. The t = 5 seconds is fine, but the t = –1 seconds doesn’t make sense, so we’ll discard it. The solution is: The rocket hits the ground 5 seconds after launch. |
m. Find out how long it takes to
hit the ground.
